ZOJ-2362 Beloved Sons 贪心

ZOJ-2362 Beloved Sons 贪心 | KM

　　题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2362
　　裸的匹配问题，直接KM，就算是O(n^4)的KM也不会超。当然注意到题目中左边的点到右点所连的边的权值是一样的，所以完全可以贪心拍个序，然后找增广路。。。
  1 //STATUS:C++_AC_250MS_848KB
  2 #include <functional>
  3 #include <algorithm>
  4 #include <iostream>
  5 //#include <ext/rope>
  6 #include <fstream>
  7 #include <sstream>
  8 #include <iomanip>
  9 #include <numeric>
 10 #include <cstring>
 11 #include <cassert>
 12 #include <cstdio>
 13 #include <string>
 14 #include <vector>
 15 #include <bitset>
 16 #include <queue>
 17 #include <stack>
 18 #include <cmath>
 19 #include <ctime>
 20 #include <list>
 21 #include <set>
 22 #include <map>
 23 using namespace std;
 24 //using namespace __gnu_cxx;
 25 //define
 26 #define pii pair<int,int>
 27 #define mem(a,b) memset(a,b,sizeof(a))
 28 #define lson l,mid,rt<<1
 29 #define rson mid+1,r,rt<<1|1
 30 #define PI acos(-1.0)
 31 //typedef
 32 //typedef __int64 LL;
 33 //typedef unsigned __int64 ULL;
 34 //const
 35 const int N=410;
 36 const int INF=0x3f3f3f3f;
 37 const int MOD=100000,STA=8000010;
 38 //const LL LNF=1LL<<60;
 39 const double EPS=1e-8;
 40 const double OO=1e15;
 41 const int dx[4]={-1,0,1,0};
 42 const int dy[4]={0,1,0,-1};
 43 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
 44 //Daily Use ...
 45 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 46 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 47 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 48 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 49 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 50 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 51 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 52 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 53 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 54 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 55 //End
 56 
 57 
 58 struct Node{
 59     int val,id;
 60     bool operator < (const Node& a)const{
 61         return val>a.val;
 62     }
 63 }nod[N];
 64 int ca;
 65 
 66 int w[N][N],y[N],vis[N];
 67 int n,m;
 68 
 69 int dfs(int u)
 70 {
 71     int v;
 72     for(v=1;v<=n;v++){
 73         if(w[u][v] && !vis[v]){
 74             vis[v]=1;
 75             if(y[v]==-1 || dfs(y[v])){
 76                 y[v]=u;
 77                 return 1;
 78             }
 79         }
 80     }
 81     return 0;
 82 }
 83 
 84 int main()
 85 {
 86  //   freopen("in.txt","r",stdin);
 87     int i,j,tot,a;
 88     int x[N];
 89     scanf("%d",&ca);
 90     while(ca--)
 91     {
 92         scanf("%d",&n);
 93         for(i=1;i<=n;i++){
 94             scanf("%d",&nod[i].val);
 95             nod[i].val*=nod[i].val;
 96             nod[i].id=i;
 97         }
 98         mem(w,0);
 99         for(i=1;i<=n;i++){
100             scanf("%d",&tot);
101             while(tot--){
102                 scanf("%d",&a);
103                 w[i][a]=nod[i].val;
104             }
105         }
106         sort(nod+1,nod+n+1);
107 
108         mem(x,0);
109         mem(y,-1);
110         for(i=1;i<=n;i++){
111             mem(vis,0);
112             dfs(nod[i].id);
113         }
114         for(i=1;i<=n;i++)
115             if(y[i]!=-1)x[y[i]]=i;
116 
117         printf("%d",x[1]);
118         for(i=2;i<=n;i++)
119             printf(" %d",x[i]);
120         putchar('
');
121     }
122     return 0;
123 }